✔ 最佳答案
a) Prove by mathematical induction that 1^(2)+2^(2)+3^(2)+...+n^(2)=1/6．ｎ(n+1)(2n+1)for all positive integers n.
let S(n) be the statement " 1^(2)+2^(2)+3^(2)+...+n^(2)=1/6．ｎ(n+1)(2n+1)for all positive integers n."
when n=1
LHS=1
RHS=1
so, when n=1 the statement is true
Assume when n=k, the statement is true that is
1^(2)+2^(2)+3^(2)+...+k^(2)=1/6*k(k+1)(2k+1) for all positive integers n.
when n=k+1
1^(2)+2^(2)+3^(2)+...+k^(2)+(k+1)^2
=1/6*k(k+1)(2k+1)+(k +1)^2
=1/6(k+1)[k(2k+1)+6k+6]
=1/6(k+1)[2k^2+7k+6]
=1/6(k+1)(k+2)(2k+3)
when n=k+1, the statement is true
By MI for all values of n, S(n) is true
b) Hence,or otherwise,find the values of
i)
2^(2)+4^(2)+6^(2)+...+(2n)^(2)
=(2^2)(1^2)+(2^2)(2^2)+(2^2)(3^2)+...+(2^2)(n^2)
=(2^2)[(1^2)+(2^2)+(3^2)+...+(n^2)]
=4*1/6(n+1)(n+2)(2n+ 3)
=2/3(n)(n+1)(2n+1)
ii)
1^(2)-2^(2)+3^(2)-4^(2)+...-40^(2)
because
(1^2)+(2^2)+(3^2)+...+(40^2)
=1/6(40*41*81)
=22140
(2^2)+(4^2)+(6^2)+...+(40^2)
=2/3(20)(21)(41)
=11480
[(1^2)+(2^2)+(3^2)+...+(40^2)]-[1^(2)-2^(2)+3^(2)-4^(2)+...-40^(2)]
=2[(2^2)+(4^2)+(6^2)+...+(40^2)]
=2*11480
=22960
1^(2)-2^(2)+3^(2)-4^(2)+...-40^(2)
=22140-22960
=-820
c) Using the result obtain in (a),or otherwise,deduce a formula for the expression
1．1+2．3+3．5+...+n(2n1)
1．1+2．3+3．5+...+n(2n1)
=Σi(2i-1) (i from 1 up to n)
=Σ(2i^2-i)
=2Σi^2-Σi
=1/3ｎ(n+1)(2n+1)-1/2n(n+1 )
=1/6n(n+1)[2(2n+1)-3]
=1/6n(n+1)(4n-1)
